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I am reading through Rudin's Principles of Mathematical Analysis and had a few related questions.

First, Rudin defines an open set, $E$, as a set such that every point is an interior point. A point is an interior point if there is a neighborhood, $N \subset E$ that contains the point. This neighborhood depends on a metric. So in this book (at least so far), open sets are only considered in metric spaces. But open sets need not depend on a metric.

Now Theorem 2.34 says that compact subsets of metric spaces are closed and Theorem 2.35 says that closed subsets of compact sets are compact. I understand and agree with both of these proofs, but was wondering a bit about Theorem 2.35. Obviously, in a metric space an open subset of a compact set is not compact by Theorem 2.34.

My question is: Under what conditions is an open subset of a compact set compact? I don't know much topology, but are there some easy examples, or is this not possible?

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    $\begingroup$ Open sets in a metric space CAN be compact. They just also have to be closed. The closed interval as a subset of itself is open, and also compact. $\endgroup$ – Callus May 26 '15 at 15:33
  • $\begingroup$ I would like to point out in this comment that the notion of an interval as a subset of itself equips it with some additional structure, which I don't know if you are familiar with, called the 'subspace topology' in which a set is open iff it is open on intersection with the subspace. So to see the closed interval as a subspace of itself is open, consider the whole space, and intersect it. The whole space is open, and so their intersection is open in the subspace topology. $\endgroup$ – Alfred Yerger May 26 '15 at 15:37
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Recall that a set is compact if and only if it is complete and totally bounded. A metric space is a Hausdorff space, so compact sets are closed. Therefore a compact open set must be both open and closed. If $X$ is a connected metric space, then the only candidates are $\varnothing$ and $X$. For example, if $X\subset\mathbb R^n$ then $X$ is open and compact (in the subspace topology) if and only if $X$ is bounded. However, if $X$ is disconnected, then proper subsets can be open and compact. For example, if $X=[0,1]\cup [2,3]$, then $[0,1]$ and $[2,3]$ are compact open sets in $X$.

Another example is any finite set $X=\{x_1,\ldots,x_n\}$ (regardless of the metric). For if $U_\alpha$ is an open cover of $X$, then for each $j$ there is $\alpha_j$ such that $x_j\in U_{\alpha_j}$ for each $j$, so $\{U_{\alpha_1}, \ldots, U_{\alpha_n}\}$ is a finite subcover.

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If your compact space has several connected components, then each of them is simultaneously open and closed; being closed subsets of a compact space, they are compact. A concrete example: the connected components of $O(n)$.

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  • $\begingroup$ What is meant by $O(n)$ here? $\endgroup$ – Ebearr May 26 '15 at 16:34
  • $\begingroup$ The orthogonal group of $\mathbb R ^n$. $\endgroup$ – Alex M. May 26 '15 at 16:41
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I can't think of any really interesting examples, but I can give one kind of trivial example. Consider a space with just a finite number of points, and let's give it the discrete topology. Then every set in this space is open, and closed. Furthermore, if you take an open cover of any set, there is a finite cover, since the one-point sets are open and there are only finitely many of them.

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